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Question

If a,b,c are three complex numbers such that a2+b2+c2=0 and
Δ=∣ ∣ ∣b2+c2abacabc2+a2bcacbca2+b2∣ ∣ ∣=ka2 b2 c2
then value of k is

A
1
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B
2
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C
2
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D
4
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Solution

The correct option is D 4
Using a2+b2+c2=0 , we can write Δ as
Δ=∣ ∣ ∣a2abacabb2bcacbcc2∣ ∣ ∣=abc∣ ∣aaabbbccc∣ ∣
[Taking a,b,c commonfrom C1,C2,C3 respectively]
a2b2c2∣ ∣111111111∣ ∣=a2b2c2∣ ∣021001201∣ ∣
[Using C1C1+C2 and C2C2+C3]
=4a2b2c2
Thus k=4

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